The two wires A and B of equal cross-section but of different materials are joined together. The ratio of Young's modulus of wire A and wire B is 20/11 . When the joined wire is kept under certain tension the

Question

Quantrex Academy · Accepted Answer

Correct answer: C. 1.21

Given that the wires have equal cross-sectional area

A

and are under the same tension

F

, the elongation

\Delta L

is given by:

\Delta L = \dfrac{F L}{A Y}

Since the elongations in both wires are equal (

\Delta L_A = \Delta L_B

):

\dfrac{F L_A}{A Y_A} = \dfrac{F L_B}{A Y_B}

\dfrac{L_A}{Y_A} = \dfrac{L_B}{Y_B}

L_B = L_A \times \dfrac{Y_B}{Y_A}

Substituting the given values

L_A = 2.2

m and

\dfrac{Y_A}{Y_B} = \dfrac{20}{11}

:

L_B = 2.2 \times \dfrac{11}{20}

L_B = \dfrac{24.2}{20} = 1.21

m

Answer:

1.21

The two wires A and B of equal cross-section but of different...

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